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Chapter 13: Problem 61

Reverse the order of integration in the following integrals. $$\int_{0}^{1} \int_{0}^{\cos ^{-1} y} f(x, y) d x d y$$

Short Answer

Expert verified
The new limits of integration when reversing the order are: $$\int_{0}^{\frac{\pi}{2}} \int_{\cos(x)}^{1} f(x, y) d y d x$$

Step by step solution

01

Understand the original limits of integration

The given integral is: $$\int_{0}^{1} \int_{0}^{\cos ^{-1} y} f(x, y) d x d y$$ These limits tell us that y varies from 0 to 1 and x varies from 0 to \(\cos^{-1}(y)\).
02

Draw the region of integration

To better understand the area being integrated, let's draw the region defined by the given limits of integration. We need to consider the following bounds: 1. y = 0 2. y = 1 3. x = 0 4. x = \(\cos^{-1}(y)\) This corresponds to a region where the lower bound is x = 0 (the vertical axis), the left bound is y = 0 (the horizontal axis), the upper bound is y = 1, and the curve x = \(\cos^{-1}(y)\) delimits the right curve which gives us a triangular shape with x-values going from 0 to \(\frac{\pi}{2}\) as y goes from 0 to 1.
03

Determine the new limits of integration while switching order

To reverse the order of integration, we need to describe the same region using new limits in terms of x and y. We want to integrate y first, then x. From the above description, we know that x goes from 0 to \(\frac{\pi}{2}\). In terms of y, we already know the inverse of x = \(\cos^{-1}(y)\) is y = \(\cos(x)\). Thus, along the x-axis, the lower bound of y is \(\cos(x)\), and its upper bound is 1. Now, we can rewrite the reversed integral as follows: $$\int_{0}^{\frac{\pi}{2}} \int_{\cos(x)}^{1} f(x, y) d y d x$$
04

Present the reversed integral

The reversed order of integration for the given integral is: $$\int_{0}^{\frac{\pi}{2}} \int_{\cos(x)}^{1} f(x, y) d y d x$$

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Most popular questions from this chapter

A disk of radius \(r\) is removed from a larger disk of radius \(R\) to form an earring (see figure). Assume the earring is a thin plate of uniform density. a. Find the center of mass of the earring in terms of \(r\) and \(R\) (Hint: Place the origin of a coordinate system either at the center of the large disk or at \(Q\); either way, the earring is symmetric about the \(x\) -axis.) b. Show that the ratio \(R / r\) such that the center of mass lies at the point \(P\) (on the edge of the inner disk) is the golden mean \((1+\sqrt{5}) / 2 \approx 1.618\)

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